How do you solve Smallest Index With Digit Sum Equal to Index on LeetCode?

Smallest Index With Digit Sum Equal to Index (LeetCode #3550) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Digit sum can be computed efficiently.

What is the time complexity of Smallest Index With Digit Sum Equal to Index?

The optimal solution for Smallest Index With Digit Sum Equal to Index runs in O(n) time and uses O(1) space. We only traverse the array once, making it O(n) in time complexity, and we use constant space.

What is the brute force solution for Smallest Index With Digit Sum Equal to Index?

The brute force approach for Smallest Index With Digit Sum Equal to Index is Brute Force, with O(n²) time complexity and O(1) space complexity. Check each index and calculate the digit sum for the corresponding number. If it matches the index, return that index. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Smallest Index With Digit Sum Equal to Index?

Smallest Index With Digit Sum Equal to Index uses the following concepts: Array, Math. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Smallest Index With Digit Sum Equal to Index?

Explain your thought process clearly. Consider edge cases like empty arrays. Optimize your solution after the brute force.

What are common mistakes when solving Smallest Index With Digit Sum Equal to Index?

Forgetting to convert numbers to strings for digit sum. Not handling the case where no index matches.

#3550

Smallest Index With Digit Sum Equal to Index

Easy

ArrayMathHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Iterate through the array once, calculating the digit sum on-the-fly and checking against the index.

⚙️

Algorithm

3 steps

1Step 1: Initialize a loop to go through each index i in nums.
2Step 2: For each nums[i], calculate the digit sum directly.
3Step 3: If the digit sum equals i, return i; if no match after the loop, return -1.

solution.py5 lines

1def smallestIndex(nums):
2    for i in range(len(nums)):
3        if sum(int(d) for d in str(nums[i])) == i:
4            return i
5    return -1

ℹ

Complexity note: We only traverse the array once, making it O(n) in time complexity, and we use constant space.

1Digit sum can be computed efficiently.
2The problem requires checking each index once.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.